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blaisepascal
I'm taking an online college-level course, MITx 6.002x Circuits and Electronics. As may be evident from the name and number, it's related to the freshman MIT 6.002 Circuits and Electronics class. No MIT credit, all online, and free (there will be a certificate of completion at the end).

Last night the 6th problem set and lab were due (at midnight). Around 10:30pm I had answered all but one problem (which in turn had about 5 parts, so it was a big problem), and realized that the problem I was hitting me right where I was weakest, that I wasn't going to get it, and that I was tired, after having driven 7 hours from Rhode Island early. So I have up on the problem, knowing from looking at the instant feedback that I was resigning to breaking my perfect score with a 62% on this assignment. Perhaps I can make it one of the two homeworks I get to drop.

But I am determined to review the solution to the problem set to see where I should have gone with it. It hasn't been released yet, but should be tonight.

But now... Because of a bug on the part of the course, the homework deadline has been extended to tomorrow night. The solutions will still be released tonight, but we are expected to honor the Honor Code and not look at the solutions before turning in the homework.

So... Do I take the 62% and read the solution, or do I try to figure out the problem again until I get it?

(For those who are interested, the problem is centered around the small-signal analysis of a two MOSFET phase inverter. I have to compute the minimum supply voltage to keep the transistors in saturation, the transconductance of the amplifier, etc, based on the K and VT of the matched MOSFETs. I apparently failed to grasp how to do small-signal analysis of a non-linear circuit when following the lectures. I will have to read the book carefully.)

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Damn you, damn you :) Now I am looking this up instead of cleaning and organizing!

To actually answer your question, it's almost finals time. How many more HW's are there, and secondly does the grade matter other than to you? I feel in this case I would gain more from learning how to do the analysis.

The online course doesn't run exactly parallel to the off-line course. It's not almost finals time; it's almost mid-term time. I've done 6 weeks of problem sets and labs, and there are 6 more weeks of problem sets and labs to go, plus a mid-term and a final.

The grade matters to no one but me. No one asked me to take this class, and it isn't directly related to my occupation so my employer doesn't care. The certificate of completion, plus $5, will get me a cup of coffee at Starbucks. And then I'd print out another copy of the certificate.

I'll re-read the text on small-circuit analysis, and try again.

This isn't the right circuit; this is a DC->AC converter, suitable for driving an AC-based appliance off of a battery. It also used 4 MOSFETs rather than 2. Since DC->AC converters are also called "inverters" (since they invert the result of a rectifier), and this inverter generates a single-phase output. Hence, a "single phase inverter"

I'm guessing, based on the circuit design, the questions, and the problem title, that the result of the circuit I have to analyze will be a linear amplifier with a gain of A=-1, so the phase of the input signal is inverted in the output signal, hence a "phase inverter".

(For specifics of the problem, it is assumed that the MOSFETs have the property of Ids = (K/2)(Vgs-VT)^2, where Ids is the current from drain to source, Vgs is the gate-source voltage, VT is the threshold voltage of the MOSFET, and K is a parameter determined by the MOSFET. The circuit consists of the two MOSFETs connected in series drain-to-source from the supply voltage Vdd to ground, with the output Vout taken from between the MOSFETs. The high-side FET's gate is connected to Vdd as well, and the low-side FET's gate is connected to the input Vin. It is assumed that K and VT are identical for both transistors. The first question in the problem is if the input Vin = VIN + vin, where VIN is a bias voltage and vin is the small voltage signal with |vin|<Dv, what is the minimum possible value of VIN to keep the transistor out of cutoff? This was easy. The second question: what's the minimum value of Vdd to keep both transistors in saturation? Stumped. They get harder from there.)

It just hit me... The current through both transistors has to be equal (there's no other place for it to go), so the current in the top transistor (Vdd-Vout-VT)^2 has to equal the current in the bottom transistor (Vin-VT)^2 (modulo a common factor of K/2). This means that Vdd-Vout has to equal Vin. Neither K nor VT factor in at all.

Since (modulo a constant offset) Vout = -Vin, it's clear that I was exactly right, it's a linear amplifier with a gain of -1, a perfect phase inverter.

Won. The 62% is now 100%.

Now onto week 7 ;-). But first, Rock Band!

Don't look at me - I often have to translate circuits into their tube equivalents to really grasp what they're doing :-(

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